Your answer would be 3.75.
We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.
First of all, we will find z-score corresponding to 38 and 56.


Now we will find z-score corresponding to 56.

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is
.
We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.
We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
Answer:

Step-by-step explanation:
The distance (d) between two points in 3 dimensions is ...
d = √((x2 -x1)² +(y2 -y1)² +(z2 -z1)²)
Then the distance between (a, -b, -4) and (0, 0, 0) is ...
d = √((a -0)² +(-b -0)² +(-4 -0)²)
= √(a² +b² +16)
5x=39 x stands for the age of bob so 39/5 =7.8 which is rounded up to 8 and if you multiply 8 by 5 you get 40 but if you multiply 7.8 by 5 you get 39 so his age would be 7.8 years old.